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Tyrianne solved a quadratic equation. Her work is shown below. In which step did Tyrianne make an error?

[tex]\[
\begin{array}{c}
\frac{1}{2}(x+4)^2-3=29 \\
\frac{1}{2}(x+4)^2=32 \\
(x+4)^2=64 \\
x+4= \pm 8 \\
x=4 \text{ or } x=-12
\end{array}
\][/tex]

A. Step 1
B. Step 2
C. Step 3
D. Step 4

Sagot :

GinnyAnswer
Let's go through each step of Tyrianne's work to identify where an error might have occurred.

1. [tex]\(\frac{1}{2}(x+4)^2 - 3 = 29\)[/tex]
This is the original equation given. There is no mistake here.

2. [tex]\(\frac{1}{2}(x+4)^2 = 32\)[/tex]
Tyrianne added 3 to both sides to isolate the quadratic term. This step is correct.

3. [tex]\((x+4)^2 = 64\)[/tex]
Tyrianne multiplied both sides by 2 to eliminate the fraction. This step is also correct.

4. [tex]\(x+4 = \pm 8\)[/tex]
Tyrianne took the square root of both sides. Taking the square root of 64 correctly gives [tex]\(\pm 8\)[/tex]. This step is correct.

5. [tex]\(x = 8 - 4 \text{ or } x = -8 - 4\)[/tex]
Therefore, the solutions should be [tex]\(x = 4 \text{ or } x = -12\)[/tex]. Tyrianne incorrectly simplified the equation to [tex]\(x = 0 \text{ or } x = -8\)[/tex] instead of [tex]\(x = 4 \text{ or } x = -12\)[/tex].

The mistake was made in Step 5.

So, Tyrianne made an error in Step 5.
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